• bostjan64
      Do you know how to integrate? The integral of y dx is the area between y(x) and y=0. If you haven't learned that expression yet, I can see how the equation would be difficult to solve.
  • I'm confident I can when I learn how! haha Wait, I'm gonna guess y=A
    • bostjan64
      Good guess, but that guess doesn't really work unless A is identically zero (it might be, but that's not specified in the equation). If A is a constant with respect to y and x, and y = A, then dy/dx (the rate of change in y as one varies x) would be zero, which cannot be true for all x unless A=0.
  • The key to solving any differential equation is to recognize which method will be easiest to use. In this case, the variables in the equation are separable, so I will solve it by separation of variables. Start by multiplying each side of the equation by dx. So dy/dx dx = just dy. A x dx is already as simple as it can be for now. Next, integrate both sides of the equation. Integrating a polynomial with one variable is easy- increase the exponent of the variable in each term by one and divide the coefficient of each term by the new exponent of the variable in that term. The indefinite integral of dy is y + a. The indefinite integral of A x dx is A/2 x^2 + b. For simplicity's sake, lets take C = b - a, so that we can simplify the solution to y = A/2 x^2 + C. Always check your work. So, if y = A/2 x^2 + c, dy/dx = 2A/2 x^1 = A x. Bingo.
    • Linda Joy
      Sure this you will explain in great detail, but I have to go look up whence frog in your throat came from?!! BTW once you turned your back to face the board I started imitating you behind your back and disrupting class! hehe

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